Differential Equations Exercise 3

In this exercise, we are dealing with a second-order differential equation:

$$ y''=\cos(x) $$

This is an elementary second-order differential equation.

To determine the unknown function, we simply need to integrate twice.

Let’s start by integrating both sides:

$$ \int y'' \, dx = \int \cos(x) \, dx $$

The integral on the right gives the antiderivative sin(x) + c₁:

$$ \int y'' \, dx = \sin(x) + c_1 $$

The integral on the left is just y′:

$$ y' = \sin(x) + c_1 $$

Next, we integrate once more:

$$ \int y' \, dx = \int \sin(x) + c_1 \, dx $$

The integral on the right yields the antiderivative -cos(x) + c₁x + c₂:

$$ \int y' \, dx = - \cos(x) + c_1 \cdot x + c_2 $$

The integral on the left gives the unknown function y:

$$ y = - \cos(x) + c_1 \cdot x + c_2 $$

This expression represents the general solution of the differential equation.

And that concludes the exercise.